%%% INTRODUCTION %%%
\section{Introduction to the Problem}
\label{introduction}
The aim of this project is to investigate the relationship between topological spaces and vector spaces. As such, we will 
be studying various topological vector spaces (TVS) including normable Hilbert and Banach spaces, other metrizable (but not
necessarily normable) spaces like Fr\'echet spaces, and finally non-metrizable spaces. We will state the main important properties about these spaces and 
cover relevant concepts in Functional Analysis and Theory of Distributions. 

\subsection{Definitions}
We will start with some definitions which will be important for the subsequent chapters.

%%%VECTOR SPACE%%%
\subsubsection{Definition}
A \textit{vector space over} $\mathbf{F}$ is a non-empty set $V$ together with two functions, one from $V \times V$ to $V$ and the other from $\mathbf{F} \times V$ to $V$, denoted by $x+y$ and $\alpha x$ respectively, for all $x, y \in V$ and $\alpha \in \mathbf{F}$, such that, for any $\alpha, \beta \in \mathbf{F}$ and any $x,y,z \in V$,
\begin{enumerate}
	\item[(a)] $x + y = y + x, \qquad x + (y + z) = (x + y) + z$;
	\item[(b)] there exists a unique $0 \in V$ (independent of $x$) such that $x+0=x$;
	\item[(c)] there exists a unique $-x \in V$ such that $x + (-x) = 0$;
	\item[(d)] $1x = x, \qquad \alpha(\beta x) = (\alpha \beta) x$;
	\item[(e)] $\alpha (x+y) = \alpha x + \alpha y, \qquad (\alpha + \beta ) x = \alpha x + \beta x$.
\end{enumerate}
 \cite{defs}

%%%TOPOLOGICAL SPACE%%%
\subsubsection{Definition}
A \textit{topology} on a set $X$ is a family $T$ of subsets of $X$ with the following properties:
\begin{enumerate}
	\item[(i)] $\emptyset , X \in T$;
	\item[(ii)] the union of the members of any subfamily of $T$ is a member of $T$;
	\item[(iii)] the intersection of a finite number of members of $T$ is a member of $T$.
\end{enumerate}
The pair $(X,T)$ is called a \textit{topological space}; the members of T are called its \textit{open set}. \cite{defs3}

\subsubsection{Definition}
A \textit{topological vector space} is a linear space $X$ that is also a topological space such that addition and scalar multiplication are continuous operations; 
that is, if $K$ is the underlying scalar field, then the mapping $(x,y) \mapsto x+y$ is continuous on $X\times X$ into $X$ and the mapping $(\alpha,x) \mapsto \alpha x$ is continuous on $K\times X$ into $X$. \cite{defs2}

%%%NEIGHBOURHOOD%%%
\subsubsection{Definition}\label{neighbourhood}
A \textit{neighbourhood} of a point $x \in X$ is any set $V \subset X$ such that there is an open set $U \subset V$ with $x \in U \subset V$. \cite{defs2}

%%%INVARIANT%%%
\subsubsection{Definition}
A space X is said to be \textit{invariant} iff
\begin{enumerate}
	\item[(i)] $T_a(x) = a+x$;
	\item[(ii)] $M_{\lambda}(x) = {\lambda}x$;
\end{enumerate}
are translation and multiplication operations which are homeomorphisms from $X$ into $X$.

%%%COMPLETE%%%
\subsubsection{Definition}
A metric space $(M,d)$ is \textit{complete} if every Cauchy sequence in $(M,d)$ is convergent. A set $A \subset M$ is \textit{complete} (in $(M,d)$) if every Cauchy sequence lying in A converges to an element of A. \cite{defs}

%%%COMPACT%%%
\subsubsection{Definition}
Let $(M,d)$ be a metric space. A set is  $A \subset M$ is \textit{compact} if every sequence $\{x_n\}$ in $A$ contains a subsequence that converges to an element of $A$. If the set $M$ itself is compact then we say that $(M,d)$ is a  \textit{compact metric space}. ~ \cite{defs}

%%%NORM%%%
\subsubsection{Definition}
Let $X$ be a vector space over $ \mathbf{F}$. A \textit{norm} on $X$ is a function  $ \| \cdot \| : X \rightarrow \mathbf{R} $ such that for all $x, y, \in X $ and $ \alpha \in \mathbf{F} $,
\begin{enumerate}
	\item[(i)] $\|x\| \geq 0$;
	\item[(ii)] $\|x\| = 0 \Leftrightarrow x = 0$;
	\item[(iii)] $\|\alpha x\| = |\alpha |\|x\|$;
	\item[(iv)] $\| x+y\| \leq \| x\| + \| y\|$.
\end{enumerate}
A vector space X on which there is a norm is called a \textit{normed vector space} or just a \textit{normed space}. ~ \cite{defs}

A topological vector space $X$ is said to be \textit{normable} if there exists a norm on $X$ such that the norm topology is the same as the given topology for $X$. ~\cite{defs2}

We will need the definition of completeness and normed vector spaces in Chapter \ref{sec:NormedSpaces} for the defining Hilbert and Banach spaces.

%%%UNIFORMLY CONTINUOUS%%%
\subsubsection{Definition}
Let $X$ and $Y$ be normed linear spaces and $T$ a linear operator $X$ into $Y$. If $T$ is continuous at some point $x_0 \in X$, then $T$ is \textit{uniformly continuous} on all of $X$. Furthermore, $T$ is (uniformly) continuous on $X$ if and only if there is a constant $M$ such that $\|Tx\| \leq M \|x\|$ for every $x$ in $X$.

\subsection{Fundemental Properties of Neighbourhoods of Zero in TVS}
As $U + x_0 = \{x + x_0 : x\in U\}$ is open iff $U$ is open, the topology of a TVS is completely determined by any 
local base (at zero). \\\\
Let $T:X\rightarrow Y$ be a linear map, where $X$ and $Y$ are TVS.
$T$ bounded $\Leftrightarrow T(A) \in Y$ is a bounded subset if $A \in X$ is bounded $\Leftrightarrow T$ bounded on some neighbouhood
of zero $\in X$ (since the subset $A$ can be mapped homeomorphically into any neighbourhood of zero). \\\\
also $T$ continuous $\Leftrightarrow T$ continuous at zero,
since we have that $$ T(x_0+U) = T(x_0) + T(U) \subset T(x_0) +V $$ where $x_0 \in X$ and $U \subseteq X$ , $V \subseteq Y$ are open sets in the respective topologies

\subsection{Continuous Linear Mappings}
\subsubsection{Definition}
If $M$ is a subspace of TVS $X$, then the \textit{quotient topology} on $X/M$ is the topology determined by the 
canonical, linear map 
$$T: X \rightarrow X/M$$
$$\qquad\qquad x \rightarrow \dot{x} = x + M$$  
 
\subsubsection{Definition}
A linear mapping $T$ from TVS $X$ into a TVS $Y$ is an (algebraic) \textit{homomorphism} if the mapping from $X/N$ 
(where $N$ is the kernel of $T$) onto $T(X)$ is a (topological) \textit{isomorphism}.
  
\subsubsection{Proposition}
Let $T$ be a continuous linear mapping from a TVS $X$ into a TVS $Y$. The following conditions are equivalent:
\begin{enumerate}
	\item[(a)] T is a (algebraic) homomorphism  
	\item[(b)] T transforms the open subsets of $X$ into open subsets of $T(X)$
	\item[(c)] T transforms the neighbourhoods of zero of $X$ into the the neighbourhoods of zero in $T(X)$
\end{enumerate}
It follows that a) $\Rightarrow$ b) $\Rightarrow$ c) $\Rightarrow$ a). The continuous linear mappings from $X$ into $Y$ then form
a TVS, denoted as $L(X,Y)$.  \cite{defs4}